Question

Factor completely: 10x2 + 39x + 14

A. (10x + 7)(x + 2)
B. (5x + 2)(2x + 7)
C. (5x + 7)(2x + 2)
D. (10x + 2)(x + 7)

Answer 1

Well, for these problems, you can actually just expand A,B,C,and D.

Answer 2

@Senior012 inkyvoyd told u a great hint :

\[\huge{10x^2+39x+140}\]
solve it through quadratic equation :
\(\huge{ax^2+bx+c=0}\) -----------------GENERAL FORM OF QUADRATIC EQUATION
\[\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

a = 10 , b = 39 , c = 140

Answer 3

solve it by putting the values of a , b and c that is
\[\huge{x=\frac{-39\pm\sqrt{(-39)^2-4(10)(140)}}{2(10)}}\]
u can solve this ..

Answer 4

may i know @Senior012 did u understand this ?

Answer 5

NO /:

Answer 6

so where r u having problem ?

Answer 7

The right answer is B
you have to break the middle term i.e 39x Into two values in such a way that the two values when added give 39 and when multiple give 140(the product of 10x2 and 14). The two values are: 35 and 4 place them in place of 39x and solve it :-)
10x2+39x+14
10x2+35x+4x+14
Take out the common from first two and then from the other two
5x(2x+7)+2(2x+7)
(5x+2)(2x+7)